Hi Guys ,
Thought this might be educative as well , the likes of Saju / Anwar / Ismail could comment and take it on further :

http://machinedesign.com/article/selecting-dc-brush-and-brushless-motors-0217

I found that understanding of the inverse relationship between speed and torque will be interesting to understand the implication of KV rating to Prop size etc.

I wonder how many have actually considered a SERVO as a Motor   so this might be of interest  to non electric modelers as well 

Sai

Why do we say a high KV Motor cannot swing a bigger prop  and why can the prop size be increased  and the KV Lowers .

Why are KV and prop size inversely proportional ?

may sound like basics to many but you will be surprised at the  burn out that occur due to these very questions.

Sai

Each motor has its own max amp rating.  So a higher KV motor will pull more amps with lesser load (smaller prop) compared to a lower KV motor spinning the same prop!! So in general on the same battery pack, if we want to spin a bigger prop, then lower KV motor is suggested and vice versa..

KV is nothing but rpm per volt at no load (correct me if I am wrong)!!

So for high speed applications like pylon racers and most of the pattern planes we use a higher KV motor with a smaller dia prop and higher pitch.

For 3D applications we use lower KV motor with bigger dia and lower pitch prop.  This would maximize the thrust required for 3D maneuvers like hovering!  However we would be compensating in speed.

Another thing is, higher KV motors put out more watts on the same voltage compared to a lower KV motor.

So if we run a 1000 kv motor and a 1800kv motor on the same 3 cell Lipo, the 1800 kv motor would be able to pull out more watts than the 1000kv motor.  But ideally, the prop size on the 1800 KV would be lesser than the 1000KV motor in order to keep the amps under the max limit.

So lets consider two planes (a 3D plane and a pattern plane) in the .50 size glow category. 

The pattern plane would take a 1800 KV motor spinning a 12x6 prop on 3 cells and would be able to keep the amps under the max limit. 

The 3D plane would take a 950 or 1000 KV motor spinning a 13x4 prop on 4 cells and still would be able to keep the amps under the max limit. 

These are just ball park figures and not very accurate.

Please let me know if this is unclear, would try to explain in different words!!

- Chan

http://www.youtube.com/watch?v=X7gJ3SvD9ks&feature=channel

Motor KV is a long subject in itself... I have not flown lot of electrics so, I cannot comment on the relation between KV and the choice of prop nor I can give any thumb rules. However, I can give a general definition of motor constants... and the theory behind why the motor coil burn on increasing the load. Pls do not curse me as the definitions are a bit too theoritical.

There are two constants for any motor :

1, Voltage constant  - KV
2, Torque constant   - KT  - Lets forget about this for a while. No need too many variables.


Voltage constant is best explanined when the motor is used as a generator. For example, when you take an ideal motor rated at 1000KV, and you spin the motor shaft at 1000 RPM you will see 1V across the motor terminals. However, in practical motors, due to losses, the voltage will be less than 1V. Note that this voltage generated by the motor is called Back Electro Motive Force (BEMF). BEMF is a vital component of the motor that opposes the applied voltage and this property helps to keep the motor coils from burning out. Further, BEMF increases with increase in RPM. Note that, if the BEMF lowers then more current will flow into the motor as there is not oppostion force from the motor to the applied voltage.

Now, to add one more confusion ---> The higher the resistance for the shaft to spin the lower the BEMF as the RPM reduces. So, now, if you have a high pitched/larger dia prop on a small motor shaft then there is more resistance for the shaft to spin (RPM reduces) and so the BEMF will be lowered. Now, more current will flow into the motor heating up the winding coils. If the motor is heavily loaded then more and more current will flow in due to reduced BEMF. Enventually the coils would burn out.

Now, back to KV and KT... The interesting thing about these two constants are that their product is always 1 for all motors

KV * KT = 1

KV = RPM / Volts
KT = Torque / Amps

Now, we have an interesting equation that applies to all motors :

RPM/Volts * Torque/Amps = 1  => Universal rule for any motor

Based on the above equation, you can decide on what could happen when one quantity increases or decreases.

-Ismail

That was a good  one Ismail .

Saju ,
You are here by invited to add your valuable comments please 

sai

Hi Rcforall

I am soon starting a thread on Understanding electric Power systems. I hope it will solve all your queries that you seem to have.

Thanks
Ankur Kaul

Quote from: ankurkaul17 on September 01, 2009, 02:30:55 PM
Hi Rcforall

I am soon starting a thread on Understanding electric Power systems. I hope it will solve all your queries that you seem to have.

Thanks
Ankur Kaul

Ankur ,
Those are not my doubts , those questions are for forum members to answer it is called interactive learning  .
I sell electrical power systems  so I know some of the answers .

All the same it would be great for you to done the hat of the teacher and we are ready to learn 

Going on from where Izmile stopped:

Torque x rpm  = Energy expended per unit time = mechanical power

Voltage x current = watts = Electric power

mechanical power / electric power = 1

i.e. mechanical power = electric power      theoretically.
This is nothing but the Law of conservation of energy


As I understand it, a high Kv rating means the speed of the prop is high for a given voltage and pitch. As the prop speed increases, the velocity of the air pushed by it increases. This energy then translates as a tube of air of cross sectional area (pi/4 x d^2 where d is the dia of the prop) moving at the velocity given to it by the prop. In other words the volume rate of air flow . If density of air is known, then the energy expended can be calculated theoretically from this. (i.e. Kinetic energy of the mass of air moving at this velocity). Putting it all together, the energy expended turns out as proportional to velocity^3 and dia ^2. Ofcourse all this is theoretical. We have to use quite a few factors for losses and efficiencies. If now the prop+motor is put on a plane with high drag (slow flyer!), the resistance for the air to flow increases and would result in increasing the torque required to rotate the prop. This would in turn start loading the motor, the esc and the battery resulting in their raise in temperature of operation. Internally, a high kv motor has a larger number of turns of copper of thin gauge, because it has to deliver a higher backemf at a lower current rating . Thus if the current increases the coil can burn if it is not sufficiently cooled

If a low kv rating is used, then the motor is only capable of delivering a lower rpm per volt. Usually, a low kv motor will have lower number of turns, but of thick wire. So the current capacity is much higher than a high kv motor. Hence it can take a higher dia prop. If the pitch of the prop is high, then the velocity of the plane has to be high, because the prop would otherwise start 'slipping' causing again an increase in torque and proportionally an increase in current. Generally if a high pitch is used, the velocity the plane is capable of should also be high. In otherwords a high pitch prop gives high speed and a low pitch prop gives low speed. A high kv rating can also result in high speed and a lower kv rating in low speed.
Thrust  x velocity = mechanical power. Increase in dia increases the thrust (because more air is being pushed by the prop)
Thrust x velocity = torque x rpm = volts x current  = power at the prop theoretically

If you are confused then I have acheived what I set out to do !!



Saju